Have you ever measured how much ice cream one cone can hold? Take a look at this dilemma.

“We need a fundraiser,” Maria said at the planning meeting for the school Olympics.

“I agree, and besides, people like to eat,” Jamie agreed.

“How about ice cream cones? We can use the freezer in the lunch room and scoop and serve,” Dan suggested.

“I think that’s a great idea. How about using waffle cones?” Maria added.

The group continued to discuss the ice cream cones and finally agreed on two different sized cones, one that is \begin{align*}4^{\prime\prime}\end{align*} in diameter and \begin{align*}4^{\prime\prime}\end{align*} long and one that is \begin{align*}5^{\prime\prime}\end{align*} in diameter and \begin{align*}6^{\prime\prime}\end{align*} long.

“We can charge double for the larger cone,” Jamie said.

“I don’t think so. It isn’t double the size,” Dan disagreed.

“But it will hold double the amount of ice cream,” Jamie explained.

“I don’t think so because it isn’t twice as large.”

“That doesn’t matter when it comes to volume,” Jamie said.

**Who is correct? To figure this out, you will need to find the volume of both cones. Then you will be able to decide whether the group can charge double for the larger cone.**

### Guidance

**What is volume?**

*Volume***is the measure of how much space a three-dimensional figure takes up or holds.**

Imagine a funnel. Its size determines how much water the funnel will hold. If we fill it with water, the amount of water tells the volume of the funnel.

Volume is often what we think of when we talk about measuring liquid or liquid capacity.

**We measure volume in three dimensions: length, width, and height. We therefore measure volume in cubic units. We can use unit cubes to represent volume.**

Cones are unusual, however, because they are so much smaller at the top than they are at their base. It becomes very difficult to use unit cubes to measure the volume of these solids because we would be calculating parts of unit cubes.

*The important thing to remember is that measuring volume involves filling up a solid figure.*

**A cone has exactly one-third the volume of a cylinder.**

Here is the formula for finding the volume of a cone.

\begin{align*}V= \frac{1}{3} Bh\end{align*}

When using this formula, you have to keep in mind that the base of a cone is circular.

**To find the base area, then, we need to use the formula for finding the area of a circle: \begin{align*}A = \pi r^2\end{align*}.**

With a cone, you will always have a circular base, so you will always be using the same formula for area to find the base. Here is what it would look like as one formula.

\begin{align*}V=\frac{1}{3} (\pi r^2)(h)\end{align*}

**Now you can see that we would find the area of the base, multiply it by the height and then multiply it by one-third or take one-third of the product of the base area and the height.**

Let’s apply this.

**What is the volume of the cone below?**

**First, we need to find the base area. The base is a circle, so we use the area formula for circles.**

\begin{align*}B & = \pi r^2\\ B & = \pi (3.5^2)\\ B & = 12.25 \pi\\ B & = 38.47 \ cm^2\end{align*}

**The circular base has an area of 38.47 square centimeters. Now we can put this measurement into the formula for volume.**

\begin{align*}V & = \frac{1}{3} Bh\\ V & = \frac{1}{3} (38.47) (22)\\ V & = 12.82 (22)\\ V & = 282.04 \ cm^3\end{align*}

**The volume of this cone is \begin{align*}282.04 \ cm^3\end{align*}.**

Do you know what a truncated cone looks like?

Notice that we have two radii to work with and the height of the truncated cone as well. We can use the following formula to calculate volume.

\begin{align*}V=\frac{1}{3} \pi(r1^2+(r1)(r2)+r2^2)h\end{align*}

*Take a few minutes and write this formula down in your notebook.*

Now we can take some measurements and figure out the volume of the truncated cone.

**What is the volume of a truncated cone with a top radius of 2 cm, a bottom radius of 4 cm and a height of 4.5 cm?**

**To work through this problem, we have to substitute the given values into the formula and solve for the volume.**

\begin{align*}V & = \frac{1}{3} \pi (r1^2+(r1)(r2)+r2^2)h \\ V & = \frac{1}{3} \pi (2^2+(2)(4)+4^2)4.5 \\ V & = \frac{1}{3} \pi (4+8+16)4.5 \\ V & = \frac{1}{3} \pi (126)\\ V & = \frac{1}{3} (395.64)\\ V & = 131.88 \ cm^3 \end{align*}

**This is the volume of this truncated cone.**

Find the volume of each cone. You may round to the nearest hundredth when necessary.

#### Example A

A cone with a radius of 2 inches and a height of 4 inches.

**Solution: \begin{align*}16.75 \ in^3\end{align*}**

#### Example B

A cone with a radius of 5 cm and a height of 7 cm.

**Solution: \begin{align*}183.17 \ cm^3\end{align*}**

#### Example C

A cone with a radius of 3 m and a height of 8 m.

**Solution: \begin{align*}75.36 \ m^3\end{align*}**

Now let's go back to the dilemma from the beginning of the Concept.

**We can begin by figuring out the volume of both ice cream cones.**

**Cone 1 has a diameter of \begin{align*}4^{\prime\prime}\end{align*} and a height of \begin{align*}4^{\prime\prime}\end{align*}**

**Cone 2 has a diameter of \begin{align*}5^{\prime\prime}\end{align*} and a height of \begin{align*}6^{\prime\prime}\end{align*}**

**The formula for volume of a cone is \begin{align*}\frac{1}{3} \pi r^2 h\end{align*}**

**Cone 1**

\begin{align*}\frac{1}{3}(3.14)(2^2)(4) = 16.75 \ in^3\end{align*}

**Cone 2**

\begin{align*}\frac{1}{3}(3.14)(2.5^2)(6) = 39.25 \ in^3\end{align*}

**The volume of Cone 2 is more than double that of Cone 1. The students could definitely charge double for the cone if they chose to.**

### Vocabulary

- Volume
- the capacity inside a solid figure or the amount of space a solid figure can hold.

- Cones
- a circular base and a curved side that meets in a single vertex.

- Base Area
- the area of the base of a solid figure.

- Height
- the measurement that is perpendicular to the base of a solid figure.

- Truncated cone
- a section of a cone-it has two circular radii – one on top and one as a base.

### Guided Practice

Here is one for you to try on your own.

**What is the height of a cone whose radius is 1.6 meters and volume is 20.1 cubic meters?**

**Solution**

What information have we been given, and what do we need to find? We know the radius, so we can calculate the base area. We also know the volume, so we can put this into the formula and solve for \begin{align*}h\end{align*}, the height. Let’s find \begin{align*}B\end{align*} first.

\begin{align*}B & = \pi r^2\\ B & = \pi (1.6^2)\\ B & = 2.56 \pi\\ B & = 8.04 \ m^2\end{align*}

**The base area is 8.04 square inches when we approximate pi. Now let’s put this into the volume formula.**

\begin{align*}V & = \frac{1}{3} Bh\\ 20.1 & = \frac{1}{3} (8.04)h\\ 20.1 & = 2.68h\\ 20.1 \div 2.68 & = h\\ 7.5 \ m & = h\end{align*}

**We found that the height of the cone must be 7.5 meters.**

### Video Review

### Practice

Directions: Answer each of the following questions.

- What is the formula for finding the volume of a cone?
- True or false. A truncated cone is a cone without a vertex.
- True or false. You can use the same formula to find the volume of a truncated cone as a regular cone.

- What is the diameter of this cone?
- What is the height of the cone?
- What is the volume of the cone?

- What is the diameter of this cone?
- What is the height of the cone?
- What is the volume of the cone?

- What is the diameter of this cone?
- What is the radius of this cone?
- What is the height of the cone?
- What is the volume of the cone?

Directions: Use what you have learned to solve each of the following problems.

- A cone has a radius of 6 meters and a volume of \begin{align*}168 \pi\end{align*}. What is its height?
- The containers of icing for Tina’s cake decorator are cones. Each container has a radius of 2.4 inches and a height of 7 inches. If Tina buys containers of red, yellow, and blue icing, how much icing will she buy?